tion list $\newline$ A group consists of four Democrats and eight Republicans. Three people are selected to attend a conference. $\newline$ a. In how many ways can three people be selected from this group of twelve? $\newline$ b. In how many ways can three Republicans be selected from the eight Republicans? $\newline$ c. Find the probability that the selected group will consist of all Republicans. $\newline$ estion $4$ $\newline$ estion $5$ $\newline$ a. The number of ways to select three people from the group of twelve is $\square$

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Q. tion list

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A group consists of four Democrats and eight Republicans. Three people are selected to attend a conference.

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a. In how many ways can three people be selected from this group of twelve?

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b. In how many ways can three Republicans be selected from the eight Republicans?

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c. Find the probability that the selected group will consist of all Republicans.

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estion

4

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estion

5

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a. The number of ways to select three people from the group of twelve is

\square

Calculate Combination Formula: To find the number of ways to select three people from the group of twelve, we use the combination formula which is $C(n, k) = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items, $k$ is the number of items to choose, and $\text{“!“}$ denotes factorial.
Calculate for Part a: For part a, we have $n = 12$ (total people) and $k = 3$ (people to select). So, we calculate rac{12!}{(3!(12-3)!)}.
Calculate for Part b: Calculating the factorials, we get $12! = 12 \times 11 \times 10 \times 9!$ , $3! = 3 \times 2 \times 1$ , and $(12-3)! = 9!$ .
Calculate for Part c: We can simplify the combination formula by canceling out the $9!$ in the numerator and denominator, which gives us $(12 \times 11 \times 10) / (3 \times 2 \times 1)$ .
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .We can simplify the combination formula by canceling out the $5!$ in the numerator and denominator, which gives us $(8 \times 7 \times 6) / (3 \times 2 \times 1)$ .
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .We can simplify the combination formula by canceling out the $5!$ in the numerator and denominator, which gives us $(8 \times 7 \times 6) / (3 \times 2 \times 1)$ .Performing the calculation, we get $220$ $0$ . So, there are $220$ $1$ ways to select three Republicans from the eight Republicans.
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .We can simplify the combination formula by canceling out the $5!$ in the numerator and denominator, which gives us $(8 \times 7 \times 6) / (3 \times 2 \times 1)$ .Performing the calculation, we get $220$ $0$ . So, there are $220$ $1$ ways to select three Republicans from the eight Republicans.For part c, to find the probability that the selected group will consist of all Republicans, we divide the number of ways to select three Republicans by the total number of ways to select three people from the group.
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .We can simplify the combination formula by canceling out the $5!$ in the numerator and denominator, which gives us $(8 \times 7 \times 6) / (3 \times 2 \times 1)$ .Performing the calculation, we get $220$ $0$ . So, there are $220$ $1$ ways to select three Republicans from the eight Republicans.For part c, to find the probability that the selected group will consist of all Republicans, we divide the number of ways to select three Republicans by the total number of ways to select three people from the group.The probability is therefore $220$ $1$ (the number of ways to select three Republicans) divided by $220$ (the total number of ways to select three people).
Find Probability: Performing the calculation, we get $(12 \times 11 \times 10) / (3 \times 2 \times 1) = 1320 / 6 = 220$ . So, there are $220$ ways to select three people from the group of twelve.For part b, we need to find the number of ways to select three Republicans from the eight Republicans. We use the same combination formula with $n = 8$ and $k = 3$ .Calculating the combination, we get $8! / (3!(8-3)!)$ .Calculating the factorials, we get $8! = 8 \times 7 \times 6 \times 5!$ , $3! = 3 \times 2 \times 1$ , and $(8-3)! = 5!$ .We can simplify the combination formula by canceling out the $5!$ in the numerator and denominator, which gives us $(8 \times 7 \times 6) / (3 \times 2 \times 1)$ .Performing the calculation, we get $220$ $0$ . So, there are $220$ $1$ ways to select three Republicans from the eight Republicans.For part c, to find the probability that the selected group will consist of all Republicans, we divide the number of ways to select three Republicans by the total number of ways to select three people from the group.The probability is therefore $220$ $1$ (the number of ways to select three Republicans) divided by $220$ (the total number of ways to select three people).Calculating the probability, we get $220$ $4$ , which can be simplified to $220$ $5$ .